Ratios and Proportions Questions

FACTS  AND  FORMULAE  FOR  RATIO  AND  PROPORTION  QUESTIONS

 

 

1. RATIO: The ratio of two quantities a and b in the same units, is the fraction a/b and we write it as a:b.

In the ratio a:b, we call a as the first term or antecedent and b, the second term or consequent.

Ex. The ratio 5: 9 represents 5/9 with antecedent = 5, consequent = 9.

Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio. Ex. 4: 5 = 8: 10 = 12: 15 etc. Also, 4: 6 = 2: 3.

 

2. PROPORTION : The equality of two ratios is called proportion

If a: b = c: d, we write, a: b :: c : d and we say that a, b, c, d are in proportion . Here a and d are called extremes, while b and c are called mean terms.

Product of means = Product of extremes.

Thus, a: b :: c : d <=> (b x c) = (a x d).

 

 3(i) Fourth Proportional : If a : b = c: d, then d is called the fourth proportional to a, b, c.

     (ii) Third Proportional : If a: b = b: c, then c is called the third proportional to a and b.

     (iii) Mean Proportional : Mean proportional between a and b is ab

 

 4. (i) COMPARISON OF RATIOS : We say that (a: b) > (c: d)  (a/b)>(c /d).

     (ii) COMPOUNDED RATIO : The compounded ratio of the ratios (a: b), (c: d), (e : f)  is    (ace: bdf)

 

5. (i) Duplicate ratio of (a : b) is a2:b2.

    (ii) Sub-duplicate ratio of (a : b) is (√a : √b).

    (iii)Triplicate ratio of (a : b) is a3:b3.

    (iv) Sub-triplicate ratio of (a : b) is a13: b13.

    (V) If ab=cd, thena+ba-b=c+dc-d (Componendo and dividendo)

 

 6. VARIATION:

(i) We say that x is directly proportional to y, if x = ky for some constant k and we write, xy

(ii) We say that x is inversely proportional to y, if xy = k for some constant k and we write, x1y

Q:

If two complimentary angles are in the ratio of 4 : 5, find the greater angle.

 

A) 40 deg B) 50 deg
C) 60 deg D) 30 deg
 
Answer & Explanation Answer: B) 50 deg

Explanation:
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1 10626
Q:

If k : l = 4 : 3 and l : m = 5 : 3, then find k : l : m ?

A) 18 : 24 : 11 B) 9 : 15 : 1
C) 20 : 15 : 9 D) 21 : 7 : 3
 
Answer & Explanation Answer: C) 20 : 15 : 9

Explanation:

Given k : l = 4 : 3
l : m = 5 : 3

Then k : l : m = 20 : 15 : 9

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12 10537
Q:

If y/(x - z) = (y + x)/z = x/y then find z : y : x ?

A) 1:2:3 B) 3:2:4
C) 4:3:2 D) 2:3:4
 
Answer & Explanation Answer: B) 3:2:4

Explanation:

Given, y/(x - z) = (y + x)/z = x/y
yz = xy + x2 - yz - xz ....(1)
Also xy = yx-z x2 -xz = y2 ....(2)
Using (1) and (2), we get yz = xy - yz + y2
2yz = xy + y2
2z = x + y
Only option (B) satisfies the equation.

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19 10401
Q:

If (2a + 3b)/(5a - 3b) = 23/5 , then the value of ba is ?

A) 5 B) 4
C) 20 D) 28
 
Answer & Explanation Answer: C) 20

Explanation:

Given (2a + 3b)/(5a - 3b) = 23/5
=> 10a + 15b = 115a - 69b
=> 105a = 84b => 5a = 4b
a/b = 4/5 => a=4 & b=5
=> bxa = 5x4 = 20.

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7 10289
Q:

Maneela lent Rs. 8000 partly at the rate of 5% and partly at the rate of 6% per annum simple interest. The total interest she get after 2 years is Rs. 820, then in which ratio will Rs. 8000 is to be divided?

A) 7:1 B) 13:5
C) 15:7 D) 2:7
 
Answer & Explanation Answer: A) 7:1

Explanation:

Maneela lent Rs. 8000 in two parts,

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24 9953
Q:

80 litres of diesel is required to travel 800 km using a 800 cc engine. If the volume of diesel required to cover a distance varies directly as the capacity of the engine, then how many litres of diesel is required to travel 1000 kms using 1200 cc engine ? 

A) 120 lit B) 140 lit
C) 110 lit D) 150 lit
 
Answer & Explanation Answer: D) 150 lit

Explanation:

To cover a distance of 800 kms using a 800 cc engine, the amount of diesel required = 80 litres.

=> the amount of diesel required to cover a distance of 1000 kms = 1000 x 80/800 = 100 litres

However, the vehicle uses a 1200 cc engine and it is given that the amount of diesel required varies directly as the engine capacity.

i.e., for instance, if the capacity of engine doubles, the diesel requirement will double too.

Therefore, with a 1200 cc engine, quantity of diesel required = 1200/800 x 100 = 150 litres.

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13 9949
Q:

Total number of students in 3 classes of a school is 333 . The number of students in class 1 and 2 are in 3:5 ratio and 2 and 3 class are 7:11 ratio . What is the strength of class that has highest number of students ?

A) 125 B) 155
C) 135 D) 165
 
Answer & Explanation Answer: D) 165

Explanation:

Ratio C1:C2 = 3:5 and C2:C3 = 7:11
So C1:C2:C3 = 21 : 35 : 55
Let the strength of three classes are 21x, 35x and 55x respectively, then
Given that 21x + 35x + 55x = 333
=> 111x = 333 or x=3
So strength of the class with highest number of students =55x = 55x3 = 165.

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10 9922
Q:

A sum is divided among K, L and M in such a way that for each rupee K gets, L gets 45 paisa and M gets 30 paisa. If the share of L is Rs. 27,  what is the total amount  ?

A) Rs. 159 B) Rs. 96
C) Rs. 147 D) Rs. 105
 
Answer & Explanation Answer: D) Rs. 105

Explanation:

K : L : M = 100 : 45 : 30
              = 20 : 9 : 6

Given L share is 27

=> 9  ------  27
      35 ------  ?

      => 105

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